POJ 1458题解

这是一道更为典型的LCS题目，不同与以往的解法，我们不用DP来做，直接用转换为LIS的思路来做。

题目：
http://poj.org/problem?id=1458
代码：

#include <cstdlib>
#include <cstdio>
#include<iostream>
#include<vector>
using namespace std;

//二分法求最长单调递增子序列关键字下标(<)
int binarySearch4(int key,int lowIndex,int highIndex,vector<int>&v2)
{
if(lowIndex==highIndex)
{
if(v2[lowIndex] == key)
{
return lowIndex;
}
else if(lowIndex == 0 && key<v2[0])
{
return lowIndex;
}
return lowIndex+1;

}
int midIndex = (lowIndex + highIndex + 1)/2;
if(key<v2[midIndex])
{
return binarySearch4(key,lowIndex,midIndex-1,v2);
}
else
{
return binarySearch4(key,midIndex,highIndex,v2);
}
}

int countTotal(vector<int>&v1)
{
int lowIndex = 0;
vector<int> v2;
v2.push_back(v1[0]);

for(int i=1;i<v1.size();i++)
{

lowIndex = binarySearch4(v1[i],0,v2.size()-1,v2);
if(lowIndex>v2.size()-1)
{
v2.push_back(v1[i]);
}
else
{
v2[lowIndex] = v1[i];
}

}
if(v2[v2.size()-1]==0)
{
return v2.size()-1;
}
else
{
return v2.size();
}
}
int main()
{
string str1;
string str2;
while(cin>>str1>>str2)
{
vector<int> v1;
//将第一个序列的元素在第二个序列的出现位置从大到小，从左到右排列起来
for(int i=0;i<str1.length();i++)
{
for(int j=str2.length()-1;j>=0;j--)
{
if(str1[i] == str2[j])
{
v1.push_back(j+1);
}
}
}
if(v1.size()==0)
{
cout<<"0"<<endl;
}
else
{
cout<<countTotal(v1)<<endl;
}
/*for(int i=0;i<v1.size();i++)
{
cout<<v1[i];
}
cout<<endl;*/
}

}

用求ＬＩＳ的方法来求ＬＣＳ的方法对应的还有一道题目：

POJ1159题：

http://poj.org/problem?id=1159

但是需要N平方的空间复杂度，因而这题A不掉，提示Memory Limit Exceeded。。。单思路应该是没错的。

对应代码：

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <vector>
using namespace std;

int n;
string str;
vector<int> v1;
vector<int> v2;

//二分法求最长单调递增子序列关键字下标(<)
int binarySearch4(int key,int lowIndex,int highIndex)
{
if (lowIndex==highIndex)
{
if (v2[lowIndex] == key)
{
return lowIndex;
}
else if (lowIndex == 0 && key<v2[0])
{
return lowIndex;
}
return lowIndex+1;

}
int midIndex = (lowIndex + highIndex + 1)/2;
if (key<v2[midIndex])
{
return binarySearch4(key,lowIndex,midIndex-1);
}
else
{
return binarySearch4(key,midIndex,highIndex);
}
}

int countTotal()
{
int lowIndex = 0;
v2.push_back(v1[0]);
//将第一个序列的元素在第二个序列的出现位置从大到小，从左到右排列起来
for (int i=1;i<v1.size();i++)
{

lowIndex = binarySearch4(v1[i],0,v2.size()-1);
if (lowIndex>v2.size()-1)
{
v2.push_back(v1[i]);
}
else
{
v2[lowIndex] = v1[i];
}

}
if (v2[v2.size()-1]==0)
{
return v2.size()-1;
}
else
{
return v2.size();
}
}

int main()
{
cin>>n>>str;
for (int i=0;i<str.length();i++)
{
for (int j=0;j<str.length();j++)
{
if (str[i] == str[j])
{
v1.push_back(str.length()-j);
}
}
}
if (v1.size()==0)
{
cout<<"0"<<endl;
}
else
{
cout<<n-countTotal()<<endl;
}
/*for (int i=0;i<v1.size();i++)
{
cout<<v1[i];
}
cout<<endl;*/
return 0;
}